THEVENINS AND NORTONS THEOREMS These are some of

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THEVENIN’S AND NORTON’S THEOREMS These are some of the most powerful analysis results to

THEVENIN’S AND NORTON’S THEOREMS These are some of the most powerful analysis results to be discussed. They permit to hide information that is not relevant and concentrate in what is important to the analysis

Low distortion audio power amplifier From Pre. Amp (voltage ) TO MATCH SPEAKERS AND

Low distortion audio power amplifier From Pre. Amp (voltage ) TO MATCH SPEAKERS AND AMPLIFIER ONE SHOULD ANALYZE THIS CIRCUIT To speakers Courtesy of M. J. Renardson http: //angelfire. com/ab 3/mjramp/index. html TO MATCH SPEAKERS AND AMPLIFIER IT IS MUCH EASIER TO CONSIDER THIS EQUIVALENT CIRCUIT! REPLACE AMPLIFIER BY SIMPLER “EQUIVALENT”

THEVENIN’S EQUIVALENCE THEOREM Thevenin Equivalent Circuit for PART A

THEVENIN’S EQUIVALENCE THEOREM Thevenin Equivalent Circuit for PART A

NORTON’S EQUIVALENCE THEOREM Norton Equivalent Circuit for PART A

NORTON’S EQUIVALENCE THEOREM Norton Equivalent Circuit for PART A

Examples of Valid and Invalid Partitions

Examples of Valid and Invalid Partitions

OUTLINE OF PROOF - version 1 If Circuit A is unchanged then the current

OUTLINE OF PROOF - version 1 If Circuit A is unchanged then the current should be the same FOR ANY Vo USE SOURCE SUPERPOSITION All independent sources set to zero in A HOW DO WE INTERPRET THIS RESULT?

OUTLINE OF PROOF - version 2 1. Because of the linearity of the models,

OUTLINE OF PROOF - version 2 1. Because of the linearity of the models, for any Part B the relationship between Vo and the current, i, has to be of the form 2. Result must hold for “every valid Part B” that we can imagine 3. If part B is an open circuit then i=0 and. . . 4. If Part B is a short circuit then Vo is zero. In this case How do we interpret this?

THEVENIN APPROACH For ANY circuit in Part B This is the Thevenin equivalent circuit

THEVENIN APPROACH For ANY circuit in Part B This is the Thevenin equivalent circuit for the circuit in Part A The voltage source is called the THEVENIN EQUIVALENT SOURCE PART A MUST BEHAVE LIKE THIS CIRCUIT The resistance is called the THEVENIN EQUIVALENT RESISTANCE

Norton Approach Norton

Norton Approach Norton

ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS Thevenin Norton This equivalence can be viewed

ANOTHER VIEW OF THEVENIN’S AND NORTON’S THEOREMS Thevenin Norton This equivalence can be viewed as a source transformation problem It shows how to convert a voltage source in series with a resistor into an equivalent current source in parallel with the resistor SOURCE TRANSFORMATION CAN BE A GOOD TOOL TO REDUCE THE COMPLEXITY OF A CIRCUIT

Source transformation is a good tool to reduce complexity in a circuit. . .

Source transformation is a good tool to reduce complexity in a circuit. . . WHEN IT CAN BE APPLIED!! “ideal sources” are not good models for real behavior of sources A real battery does not produce infinite current when short-circuited Source Transformationcan be used to determine the Thevenin or Norton Equivalent. . . BUT THERE MAY BE MORE EFFICIENT TECHNIQUES

EXAMPLE: SOLVE BY SOURCE TRANSFORMATION In between the terminals we connect a current source

EXAMPLE: SOLVE BY SOURCE TRANSFORMATION In between the terminals we connect a current source and a resistance in parallel The equivalent current source will have the value 12 V/3 k The 3 k and the 6 k resistors now are in parallel and can be combined In between the terminals we connect a voltage source in series with the resistor The equivalent source has value 4 m. A*2 k The 2 k and the 2 k resistor become connected in series and can be combined After the transformation the sources can be combined The equivalent current source has value 8 V/4 k and the combined current source has value 4 m. A Options at this point 1. Do another source transformation and get a single loop circuit 2. Use current divider to compute I_0 and then compute V_0 using Ohm’s law

PROBLEM Compute V_0 using source transformation EQUIVALENT CIRCUITS Or one more source transformation 3

PROBLEM Compute V_0 using source transformation EQUIVALENT CIRCUITS Or one more source transformation 3 current sources in parallel and three resistors in parallel

RECAP OF SOURCE TRANSFORMATION Source Transformationcan be used to determine the Thevenin or Norton

RECAP OF SOURCE TRANSFORMATION Source Transformationcan be used to determine the Thevenin or Norton Equivalent. . . WE NOW REVIEW SEVERAL EFFICIENT APPROACHES TO DETERMINE THEVENIN OR NORTON EQUIVALENT CIRCUITS

A General Procedure to Determine the Thevenin Equivalent One circuit problem 1. Determine the

A General Procedure to Determine the Thevenin Equivalent One circuit problem 1. Determine the Thevenin equivalent source Remove part B and compute the OPEN CIRCUIT voltage Second circuit problem 2. Determine the SHORT CIRCUIT current Remove part B and compute the SHORT CIRCUIT current

AN EXAMPLE OF DETERMINING THEVENIN EQUIVALENT Part B is irrelevant. The voltage V_ab will

AN EXAMPLE OF DETERMINING THEVENIN EQUIVALENT Part B is irrelevant. The voltage V_ab will be the value of the Thevenin equivalent source. What is an efficient technique to compute the open circuit voltage? Now for the short circuit current Lets try source superposition NODE ANALYSIS When the current source is open the current through the short circuit is When the voltage source is set to zero, the current through the short circuit is To compute the Thevenin resistance we use For this case the Thevenin resistance can be computed as the resistance from a - b when all independent sources have been set to zero Is this a general result?

Determining the Thevenin Equivalent in Circuits with Only INDEPENDENT SOURCES Thevenin Equivalent Source is

Determining the Thevenin Equivalent in Circuits with Only INDEPENDENT SOURCES Thevenin Equivalent Source is computed as the open loop voltage Thevenin Equivalent Resistance CAN BE COMPUTED by setting to zero all the sources and then determining the resistance seen from the terminals where the equivalent will be placed “Part B” Since the evaluation of the Thevenin equivalent can be very simple, we can add it to our toolkit for the solution of circuits!! “Part B”

LEARNING BY DOING “PART B”

LEARNING BY DOING “PART B”

LEARNING EXAMPLE COMPUTE Vo USING THEVENIN In the region shown, one could use source

LEARNING EXAMPLE COMPUTE Vo USING THEVENIN In the region shown, one could use source transformation twice and reduce that part to a single source with a resistor. . Or we can apply Thevenin Equivalence to that part (viewed as “Part A”) The original circuit becomes. . . For the open loop voltage the part outside the region is eliminated And one can apply Thevenin one more time! For open loop voltage use KVL . . . and we have a simple voltage divider!!

Or we can use Thevenin only once to get a voltage divider “Part B”

Or we can use Thevenin only once to get a voltage divider “Part B” For the Thevenin voltage we have to analyze the following circuit METHOD? ? For the Thevenin resistance Source superposition, for example Contribution of the voltage source Contribution of the current source Thevenin Equivalent of “Part A” Simple Voltage Divider

LEARNING EXAMPLE USE THEVENIN TO COMPUTE Vo You have the choice on the way

LEARNING EXAMPLE USE THEVENIN TO COMPUTE Vo You have the choice on the way to partition the circuit. Make “Part A” as simple as possible Since there are only independent sources, for the Thevenin resistance we set to zero all sources and determine the equivalent resistance “Part B” For the open circuit voltage we analyze the following circuit (“Part A”). . . The circuit becomes. . .

LEARNING EXTENSION: USE THEVENIN TO COMPUTE Vo “PART B” RESULTING EQUIVALENT CIRCUIT

LEARNING EXTENSION: USE THEVENIN TO COMPUTE Vo “PART B” RESULTING EQUIVALENT CIRCUIT

LEARNING EXTENSION: COMPUTE Vo USING NORTON PART B COMPUTE Vo USING THEVENIN PART B

LEARNING EXTENSION: COMPUTE Vo USING NORTON PART B COMPUTE Vo USING THEVENIN PART B

SAMPLE PROBLEM Equivalent Resistance: Independent sources only KVL Equivalent Voltage: Node, loop, superposition… Do

SAMPLE PROBLEM Equivalent Resistance: Independent sources only KVL Equivalent Voltage: Node, loop, superposition… Do loops How about source superposition? Opening the current source: Short circuiting the voltage source This is what we need to get KVL

SAMPLE PROLEM , , , An to compute Equivalent Source. . . All independent

SAMPLE PROLEM , , , An to compute Equivalent Source. . . All independent sources The circuit can be simplified SOURCE TRANSFORMATION Voltage divider All resistors are in parallel!!

THEVENIN EQUIVALENT FOR CIRCUITS WITH ONLY DEPENDENT SOURCES A circuit with only dependent sources

THEVENIN EQUIVALENT FOR CIRCUITS WITH ONLY DEPENDENT SOURCES A circuit with only dependent sources cannot self start. (actually that statement has to be qualified a bit. What happens if ) FOR ANY PROPERLY DESIGNED CICUIT WITH ONLY DEPENDENT SOURCES This is a big simplification!! But we need a special approach for the computation of the Thevenin equivalent resistance Since the circuit cannot self start we need to probe it with an external source The source can be either a voltage source or a current source and its value can be chosen arbitrarily! Which one to choose is often determined by the simplicity of the resulting circuit

IF WE CHOOSE A VOLTAGE PROBE. . . WE MUST COMPUTE CURRENT SUPPLIED BY

IF WE CHOOSE A VOLTAGE PROBE. . . WE MUST COMPUTE CURRENT SUPPLIED BY PROBE SOURCE The value chosen for the probe voltage is irrelevant. Oftentimes we simply set it to one

IF WE CHOOSE A CURRENT SOURCE PROBE We must compute the node voltage V_p

IF WE CHOOSE A CURRENT SOURCE PROBE We must compute the node voltage V_p The value of the probe current is irrelevant. For simplicity it is often choosen as one.

LEARNING EXAMPLE FIND THEVENIN EQUIVALENT Controlling variable: Do we use current probe or voltage

LEARNING EXAMPLE FIND THEVENIN EQUIVALENT Controlling variable: Do we use current probe or voltage probe? If we use voltage probe there is only one node not connected through source Using voltage probe. Must compute current supplied

LEARNING EXAMPLE Find the Thevenin Equivalent circuit at A - B Only dependent sources.

LEARNING EXAMPLE Find the Thevenin Equivalent circuit at A - B Only dependent sources. Hence V_th = 0 To compute the equivalent resistance we must apply an external probe We choose to apply a current probe @V_1 @V_2 Controlling variable “Conventional” circuit with dependent sources - use node analysis Thevenin equivalent

SAMPLE PROBLEM Thevenin equivalent The resistance is numerically equal to V_p but with units

SAMPLE PROBLEM Thevenin equivalent The resistance is numerically equal to V_p but with units of KOhm Loop analysis Controlling variable Voltage across current probe

Thevenin Equivalent Circuits with both Dependent and Independent Sources We will compute open circuit

Thevenin Equivalent Circuits with both Dependent and Independent Sources We will compute open circuit voltage and short circuit current For each determination of a Thevenin equivalent we will solve two circuits Any and all the techniques discussed should be readily available; e. g. , KCL, KVL, combination series/parallel, node, loop analysis, source superposition, source transformation, homogeneity The approach of setting to zero all sources and then combining resistances to determine the Thevenin resistance is in general not applicable!!

Guidelines to partition: EXAMPLE Use Thevenin to determine Vo “Part A” should be as

Guidelines to partition: EXAMPLE Use Thevenin to determine Vo “Part A” should be as simple as possible. “Part B” After “Part A” is replaced by the Thevenin equivalent we should have a very simple circuit The dependent sources and their controlling variables must remain together Open circuit voltage Options? ? ? Constraint at super node KCL at super node Equation for controlling variable Solve Short circuit current Solution to the problem Negative resistances for some “a’s” Setting all sources to zero and combining resistances will yield an incorrect value!!!!

Find Vo using Thevenin Open circuit voltage Method? ? ? Super node KVL Short

Find Vo using Thevenin Open circuit voltage Method? ? ? Super node KVL Short Circuit Current Controlling variable KVL The equivalent circuit KCL The equivalent resistance cannot be obtained by short circuiting the sources and determining the resistance of the resulting interconnection of resistors

EXAMPLE: Use Thevenin to compute Vo Select your partition Now compute V_0 using the

EXAMPLE: Use Thevenin to compute Vo Select your partition Now compute V_0 using the Thevenin equivalent “Part B” Open Circuit Voltage Use loops DON’T PANIC!! Loop equations Controlling variable KVL for V_oc Loop equations Short circuit current Controlling variable Same as before Thevenin resistance

EXAMPLE The alternative for mixed sources Open circuit voltage Short circuit current Equivalent Resistance

EXAMPLE The alternative for mixed sources Open circuit voltage Short circuit current Equivalent Resistance

SAMPLE PROBLEM Mixed sources. Must compute Voc and Isc supernode Open circuit voltage KCL

SAMPLE PROBLEM Mixed sources. Must compute Voc and Isc supernode Open circuit voltage KCL at super node The two 4 k resistors are in parallel KCL at supernode Short circuit current KVL FINAL ANSWER

SAMPLE PROBLEM Mixed sources! Must compute open loop voltage and short circuit current Open

SAMPLE PROBLEM Mixed sources! Must compute open loop voltage and short circuit current Open circuit voltage For Vx use voltage divider For Vb use KVL Short circuit current Single node We need to compute V_x KCL@Vx KCL again can give the short circuit current FINAL ANSWER

LEARNING EXAMPLE DATA TO BE PLOTTED Using EXCEL to generate and plot data

LEARNING EXAMPLE DATA TO BE PLOTTED Using EXCEL to generate and plot data

LEARNING EXAMPLE DATA TO BE PLOTTED Using MATLAB to generate and plot data »

LEARNING EXAMPLE DATA TO BE PLOTTED Using MATLAB to generate and plot data » » » » Rx=[0: 0. 1: 10]'; %define the range of resistors to use Voc=12 -6*Rx. /(Rx+4); %the formula for Voc. Notice ". /" Rth=4*Rx. /(4+Rx); %formula for Thevenin resistance. plot(Rx, Voc, 'bo', Rx, Rth, 'md') title('USING MATLAB'), %proper graphing tools grid, xlabel('Rx(k. Ohm)'), ylabel('Volts/k. Ohms') legend('Voc[V]', 'Rth[k. Ohm]')

A MORE GENERAL VIEW OF THEVENIN THEOREM USUAL INTERPRETATION THIS INTERPRETATION APPLIES EVEN WHEN

A MORE GENERAL VIEW OF THEVENIN THEOREM USUAL INTERPRETATION THIS INTERPRETATION APPLIES EVEN WHEN THE PASSIVE ELEMENTS INCLUDE INDUCTORS AND CAPACITORS Thevenin

MAXIMUM POWER TRANSFER Courtesy of M. J. Renardson http: //angelfire. com/ab 3/mjramp/index. html From

MAXIMUM POWER TRANSFER Courtesy of M. J. Renardson http: //angelfire. com/ab 3/mjramp/index. html From Pre. Amp (voltage ) To speakers The simplest model for a speaker is a resistance. . . BASIC MODEL FOR THE ANALYSIS OF POWER TRANSFER

MAXIMUM POWER TRANSFER For every choice of R_L we have a different power. How

MAXIMUM POWER TRANSFER For every choice of R_L we have a different power. How do we find the maximum value? Consider P_L as a function of R_L and find the maximum of such function 3 Technically we need to verify that it is indeed a maximum Set the derivative to zero to find extreme points. For this case we need to set to zero the numerator The maximum power transfer theorem The value of the maximum The load that maximizes the power transfer for a circuit is power that can be equal to the Thevenin equivalent resistance of the circuit. transferred is ONLY IN THIS CASE WE NEED TO COMPUTE THEVENIN VOLTAGE

LEARNING EXAMPLE a We need to find the Thevenin resistance at a - b.

LEARNING EXAMPLE a We need to find the Thevenin resistance at a - b. The circuit contains only independent sources. . b Resistance for maximum power transfer If we MUST find the value of the power that can be transferred THEN we need the Thevenin voltage!!!

LEARNING EXAMPLE c d This is a mixed sources problem a 1. Find the

LEARNING EXAMPLE c d This is a mixed sources problem a 1. Find the Thevenin equivalent at a - b 2. Remember that for maximum power transfer b . . And it is simpler if we do Thevenin at c - d and account for the 4 k at the end Controlling variable: Now the short circuit current Remember now where the partition was made